How Is the Residue Calculated for 1/(z^2+4)^2 at z=2i?

Click For Summary

Discussion Overview

The discussion focuses on calculating the residue of the function \( \frac{1}{(z^2+4)^2} \) at the point \( z=2i \). Participants explore the nature of the pole at this point and the application of residue calculation techniques, particularly using the Laurent series.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how the residue is determined, noting that \( (2i)^2 = -4 \) leads to a zero denominator.
  • Another participant identifies \( z=2i \) as a pole of order 2 and provides a formula for calculating the residue using the limit of a derivative.
  • There is a query regarding whether the derivative should be taken with respect to \( z \) or \( z^2 \), leading to a clarification that the notation must be consistent.
  • A participant asks why the limit is taken as \( z \) approaches \( 2i \) instead of \( -2i \), suggesting a connection to the definition of the pole and the context of the calculation.
  • The response indicates that the choice is based on the specific pole being evaluated, referencing an external source for justification.
  • Another participant expresses understanding after the clarifications, indicating that the discussion has helped clarify their queries.

Areas of Agreement / Disagreement

Participants generally agree on the nature of the pole and the method for calculating the residue, but there are some points of contention regarding the specifics of the derivative notation and the choice of limit point.

Contextual Notes

There are unresolved questions about the notation used in the derivative and the implications of the limit choice, which may depend on the definitions and context of the residue calculation.

Physics news on Phys.org
bugatti79 said:
Folks,

I am trying to understand calculating residues.

http://www.wolframalpha.com/input/?i=residue+of+1/(z^2+4)^2+at+z=2i

How is that answer determined? I mean (2i)^2=-4 and hence denominator is 0...?
Thanks



The point [itex]\,\,z=2i\,\,[/itex] is a pole of order 2 of [itex]\,\,\frac{1}{(z^2+4)^2}=\frac{1}{(x-2i)^2(z+2i)^2}\,\,[/itex] . Thus, from the well-known

formula that stems from Laurent series, the residue is given by
[tex]\lim_{z\to 2i}\frac{1}{(2-1)!}\frac{d}{dz^2}\left(\frac{(z-2i)^2}{(z^2+4)^2}\right)=\lim_{z\to 2i}-\frac{2}{(z+2i)^3}=-\frac{i}{32}[/tex]

DonAntonio
 
DonAntonio said:
The point [itex]\,\,z=2i\,\,[/itex] is a pole of order 2 of [itex]\,\,\frac{1}{(z^2+4)^2}=\frac{1}{(x-2i)^2(z+2i)^2}\,\,[/itex] . Thus, from the well-known

formula that stems from Laurent series, the residue is given by
[tex]\lim_{z\to 2i}\frac{1}{(2-1)!}\frac{d}{dz^2}\left(\frac{(z-2i)^2}{(z^2+4)^2}\right)=\lim_{z\to 2i}-\frac{2}{(z+2i)^3}=-\frac{i}{32}[/tex]

DonAntonio

Very good, thank you. Just have 2 queries

1) shouldn't ##/dz^2## be just ##/dz##?

2) Why did you choose pole ##z-2i## instead of ##z+2i## for the limit..?
Is it to do with ##0 < |z-z_0|<R##

Thanks
 
bugatti79 said:
Very good, thank you. Just have 2 queries

1) shouldn't ##/dz^2## be just ##/dz##?



*** Of course, thanx. The "n" in [itex]\,\,\frac{1}{(n-1)!}\,\,and\,\,\frac{d^n}{dz^n}\,\,[/itex] must be the same


2) Why did you choose pole ##z-2i## instead of ##z+2i## for the limit..?



*** Because this is what they do in that link to Wolfram you gave: the pole at [itex]\,\,z=2i\,\,[/itex], and thus the limit

must be taken as z approaches this point. ***


Is it to do with ##0 < |z-z_0|<R##


*** No, and I don't have much of an idea what you mean by this within this context.

DonAntonio


Thanks

...
 
DonAntonio said:
...

Ok that makes sense, thank you.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
13
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K